Drawing 3 balls from an urn












0












$begingroup$


There are $20$ balls in an urn: $18$ red, $2$ green.



We have to draw $3$ balls and they can be put back (replaceable), so the sample space is $20^3$, or $8000$.




  1. What is the chance of picking exactly 1 green?


  2. What is the chance of picking exactly $2$ green?



I don't understand how to do these.



Edit: For question 1, I tried doing $18cdot 18cdot 2.$






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What have you tried so far?
    $endgroup$
    – platty
    Dec 8 '18 at 21:53










  • $begingroup$
    @platty I edited my question to show what I had tried
    $endgroup$
    – aryamank
    Dec 8 '18 at 21:58


















0












$begingroup$


There are $20$ balls in an urn: $18$ red, $2$ green.



We have to draw $3$ balls and they can be put back (replaceable), so the sample space is $20^3$, or $8000$.




  1. What is the chance of picking exactly 1 green?


  2. What is the chance of picking exactly $2$ green?



I don't understand how to do these.



Edit: For question 1, I tried doing $18cdot 18cdot 2.$






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What have you tried so far?
    $endgroup$
    – platty
    Dec 8 '18 at 21:53










  • $begingroup$
    @platty I edited my question to show what I had tried
    $endgroup$
    – aryamank
    Dec 8 '18 at 21:58
















0












0








0





$begingroup$


There are $20$ balls in an urn: $18$ red, $2$ green.



We have to draw $3$ balls and they can be put back (replaceable), so the sample space is $20^3$, or $8000$.




  1. What is the chance of picking exactly 1 green?


  2. What is the chance of picking exactly $2$ green?



I don't understand how to do these.



Edit: For question 1, I tried doing $18cdot 18cdot 2.$






probability







share|cite|improve this question











$endgroup$




There are $20$ balls in an urn: $18$ red, $2$ green.



We have to draw $3$ balls and they can be put back (replaceable), so the sample space is $20^3$, or $8000$.




  1. What is the chance of picking exactly 1 green?


  2. What is the chance of picking exactly $2$ green?



I don't understand how to do these.



Edit: For question 1, I tried doing $18cdot 18cdot 2.$






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 22:02









Gaby Alfonso

839316




839316










asked Dec 8 '18 at 21:51









aryamankaryamank

466




466








  • 1




    $begingroup$
    What have you tried so far?
    $endgroup$
    – platty
    Dec 8 '18 at 21:53










  • $begingroup$
    @platty I edited my question to show what I had tried
    $endgroup$
    – aryamank
    Dec 8 '18 at 21:58
















  • 1




    $begingroup$
    What have you tried so far?
    $endgroup$
    – platty
    Dec 8 '18 at 21:53










  • $begingroup$
    @platty I edited my question to show what I had tried
    $endgroup$
    – aryamank
    Dec 8 '18 at 21:58










1




1




$begingroup$
What have you tried so far?
$endgroup$
– platty
Dec 8 '18 at 21:53




$begingroup$
What have you tried so far?
$endgroup$
– platty
Dec 8 '18 at 21:53












$begingroup$
@platty I edited my question to show what I had tried
$endgroup$
– aryamank
Dec 8 '18 at 21:58






$begingroup$
@platty I edited my question to show what I had tried
$endgroup$
– aryamank
Dec 8 '18 at 21:58












1 Answer
1






active

oldest

votes


















1












$begingroup$

Your logic is close; but you need to consider the order the balls are drawn. Since the single green ball can come first, second, or third, you have to choose where to place it when constructing a triple; this gives $3(18)(18)(2)$ as the number of such ways to get exactly one green ball. Dividing this by the total number of ways to draw 3 balls gives $frac{3(18)(18)(2)}{20^3} = frac{243}{1000}$.



Now, can you use the same method to figure the chance of getting exactly two green balls?






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks, that makes sense. Using that logic, the second answer would be 3(18*2*2)/20^3.
    $endgroup$
    – aryamank
    Dec 8 '18 at 22:22










  • $begingroup$
    I also have one more question about dealing with a 'without replacement' scenario. Why is the sample space C(20,3) and not 20*19*18? Thanks!
    $endgroup$
    – aryamank
    Dec 8 '18 at 22:57










  • $begingroup$
    That depends on how you approach the question — what are the elements of your sample space? It is viable to approach it either way, you just have to count differently (in the second case, order matters, while in the first case, it does not).
    $endgroup$
    – platty
    Dec 8 '18 at 22:58










  • $begingroup$
    It is the same scenario wherein the urn has 18 red balls and 2 green balls.
    $endgroup$
    – aryamank
    Dec 8 '18 at 23:17











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

Your logic is close; but you need to consider the order the balls are drawn. Since the single green ball can come first, second, or third, you have to choose where to place it when constructing a triple; this gives $3(18)(18)(2)$ as the number of such ways to get exactly one green ball. Dividing this by the total number of ways to draw 3 balls gives $frac{3(18)(18)(2)}{20^3} = frac{243}{1000}$.



Now, can you use the same method to figure the chance of getting exactly two green balls?






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks, that makes sense. Using that logic, the second answer would be 3(18*2*2)/20^3.
    $endgroup$
    – aryamank
    Dec 8 '18 at 22:22










  • $begingroup$
    I also have one more question about dealing with a 'without replacement' scenario. Why is the sample space C(20,3) and not 20*19*18? Thanks!
    $endgroup$
    – aryamank
    Dec 8 '18 at 22:57










  • $begingroup$
    That depends on how you approach the question — what are the elements of your sample space? It is viable to approach it either way, you just have to count differently (in the second case, order matters, while in the first case, it does not).
    $endgroup$
    – platty
    Dec 8 '18 at 22:58










  • $begingroup$
    It is the same scenario wherein the urn has 18 red balls and 2 green balls.
    $endgroup$
    – aryamank
    Dec 8 '18 at 23:17
















1












$begingroup$

Your logic is close; but you need to consider the order the balls are drawn. Since the single green ball can come first, second, or third, you have to choose where to place it when constructing a triple; this gives $3(18)(18)(2)$ as the number of such ways to get exactly one green ball. Dividing this by the total number of ways to draw 3 balls gives $frac{3(18)(18)(2)}{20^3} = frac{243}{1000}$.



Now, can you use the same method to figure the chance of getting exactly two green balls?






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks, that makes sense. Using that logic, the second answer would be 3(18*2*2)/20^3.
    $endgroup$
    – aryamank
    Dec 8 '18 at 22:22










  • $begingroup$
    I also have one more question about dealing with a 'without replacement' scenario. Why is the sample space C(20,3) and not 20*19*18? Thanks!
    $endgroup$
    – aryamank
    Dec 8 '18 at 22:57










  • $begingroup$
    That depends on how you approach the question — what are the elements of your sample space? It is viable to approach it either way, you just have to count differently (in the second case, order matters, while in the first case, it does not).
    $endgroup$
    – platty
    Dec 8 '18 at 22:58










  • $begingroup$
    It is the same scenario wherein the urn has 18 red balls and 2 green balls.
    $endgroup$
    – aryamank
    Dec 8 '18 at 23:17














1












1








1





$begingroup$

Your logic is close; but you need to consider the order the balls are drawn. Since the single green ball can come first, second, or third, you have to choose where to place it when constructing a triple; this gives $3(18)(18)(2)$ as the number of such ways to get exactly one green ball. Dividing this by the total number of ways to draw 3 balls gives $frac{3(18)(18)(2)}{20^3} = frac{243}{1000}$.



Now, can you use the same method to figure the chance of getting exactly two green balls?






share|cite|improve this answer









$endgroup$



Your logic is close; but you need to consider the order the balls are drawn. Since the single green ball can come first, second, or third, you have to choose where to place it when constructing a triple; this gives $3(18)(18)(2)$ as the number of such ways to get exactly one green ball. Dividing this by the total number of ways to draw 3 balls gives $frac{3(18)(18)(2)}{20^3} = frac{243}{1000}$.



Now, can you use the same method to figure the chance of getting exactly two green balls?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '18 at 22:04









plattyplatty

3,370320




3,370320








  • 1




    $begingroup$
    Thanks, that makes sense. Using that logic, the second answer would be 3(18*2*2)/20^3.
    $endgroup$
    – aryamank
    Dec 8 '18 at 22:22










  • $begingroup$
    I also have one more question about dealing with a 'without replacement' scenario. Why is the sample space C(20,3) and not 20*19*18? Thanks!
    $endgroup$
    – aryamank
    Dec 8 '18 at 22:57










  • $begingroup$
    That depends on how you approach the question — what are the elements of your sample space? It is viable to approach it either way, you just have to count differently (in the second case, order matters, while in the first case, it does not).
    $endgroup$
    – platty
    Dec 8 '18 at 22:58










  • $begingroup$
    It is the same scenario wherein the urn has 18 red balls and 2 green balls.
    $endgroup$
    – aryamank
    Dec 8 '18 at 23:17














  • 1




    $begingroup$
    Thanks, that makes sense. Using that logic, the second answer would be 3(18*2*2)/20^3.
    $endgroup$
    – aryamank
    Dec 8 '18 at 22:22










  • $begingroup$
    I also have one more question about dealing with a 'without replacement' scenario. Why is the sample space C(20,3) and not 20*19*18? Thanks!
    $endgroup$
    – aryamank
    Dec 8 '18 at 22:57










  • $begingroup$
    That depends on how you approach the question — what are the elements of your sample space? It is viable to approach it either way, you just have to count differently (in the second case, order matters, while in the first case, it does not).
    $endgroup$
    – platty
    Dec 8 '18 at 22:58










  • $begingroup$
    It is the same scenario wherein the urn has 18 red balls and 2 green balls.
    $endgroup$
    – aryamank
    Dec 8 '18 at 23:17








1




1




$begingroup$
Thanks, that makes sense. Using that logic, the second answer would be 3(18*2*2)/20^3.
$endgroup$
– aryamank
Dec 8 '18 at 22:22




$begingroup$
Thanks, that makes sense. Using that logic, the second answer would be 3(18*2*2)/20^3.
$endgroup$
– aryamank
Dec 8 '18 at 22:22












$begingroup$
I also have one more question about dealing with a 'without replacement' scenario. Why is the sample space C(20,3) and not 20*19*18? Thanks!
$endgroup$
– aryamank
Dec 8 '18 at 22:57




$begingroup$
I also have one more question about dealing with a 'without replacement' scenario. Why is the sample space C(20,3) and not 20*19*18? Thanks!
$endgroup$
– aryamank
Dec 8 '18 at 22:57












$begingroup$
That depends on how you approach the question — what are the elements of your sample space? It is viable to approach it either way, you just have to count differently (in the second case, order matters, while in the first case, it does not).
$endgroup$
– platty
Dec 8 '18 at 22:58




$begingroup$
That depends on how you approach the question — what are the elements of your sample space? It is viable to approach it either way, you just have to count differently (in the second case, order matters, while in the first case, it does not).
$endgroup$
– platty
Dec 8 '18 at 22:58












$begingroup$
It is the same scenario wherein the urn has 18 red balls and 2 green balls.
$endgroup$
– aryamank
Dec 8 '18 at 23:17




$begingroup$
It is the same scenario wherein the urn has 18 red balls and 2 green balls.
$endgroup$
– aryamank
Dec 8 '18 at 23:17


















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